Video Transcript
In this video, we’re going to learn
about velocity vectors: What they are? What information they show us? And how we can combine them
properly?
Imagine that you are an airplane
pilot and your work takes you to various islands throughout the Pacific Ocean. After learning about a shipwreck on
one particular island, you fly a mission to drop vital food and supplies to the
survivors on the island. As you head due east, the island is
to the northeast of your current bearing. As you fly along, a steady wind out
of the north begins to affect your flight path. You want to figure out, given this
wind, what your velocity vector should be so that you head straight towards the
island and you are able to get there as soon as possible.
To figure out this direction, we’ll
need to understand velocity vectors. When we talk about velocity
vectors, we’re talking about something that shows an object’s instantaneous speed as
well as its direction. Imagine we had a three-dimensional
coordinate space, where our dimensions were velocity in the 𝑥-direction, velocity
in the 𝑦-direction, and velocity in the 𝑧-direction. If we picked a particular point in
that space, it would have components: 𝑣 sub 𝑥, 𝑣 sub 𝑦, and 𝑣 sub 𝑧. If we wanted to write that point as
a velocity vector, we would write that it is equal to 𝑣 sub 𝑥 in the 𝑖 direction,
𝑣 sub 𝑦 in the 𝑗, and 𝑣 sub 𝑧 in the 𝑘.
We’ve said that a velocity vector
shows two things: an object’s speed and its direction. In terms of direction, that’s given
by the unit vectors 𝑖, 𝑗, and 𝑘 in this expression. The object’s speed is given by the
magnitude of the velocity vector — that is, its length. So if we have the components of the
velocity 𝑣, we can insert those, square them, add them together, and take their
square root to find speed. That’s a bit about individual
velocity vectors. Now, let’s consider what will
happen if we add them together.
In general, it’s the case that
vectors can be added together if they have the same units. For example, if a walker and a
runner were approaching one another at different velocities and we called the
walker’s velocity of one meter per second to the right 𝑣 sub one and the runner’s
velocity of two meters per second to the left 𝑣 sub two. Then we could solve for a total
velocity by adding 𝑣 sub one and 𝑣 sub two. Because these two vectors have the
same units, meters per second, we can combine them and find, if we want to
graphically, the resultant total velocity vector. Let’s get some practice adding
velocity vectors together through an example.
An athlete crosses a
30.0-meter-wide river by swimming perpendicular to the water current at a speed of
0.500 meters per second relative to the water. He reaches the opposite side at a
distance 75.0 meters downstream from his starting point. How fast is the water in the river
flowing with respect to the ground? What is the speed of the swimmer
with respect to a friend at rest on the bank of the river?
We want to solve in this exercise
for two things: first, the speed of the water in the river with respect to the
ground. We’ll call that 𝑣 sub 𝑤. Second, we want to solve for the
speed of the swimmer with respect to a stationary friend on the bank. We’ll call that speed 𝑣 sub
𝑠. We’re told the river is
3.0 [30.0] meters wide and that the swimmer swims at 0.500 meters per
second relative to the water and also that he ends up at distance 75.0 meters
downstream from his starting point. Let’s draw a diagram of this
situation to start our solution.
We have a river that’s 30.0 meters
wide. We can call that width 𝑤. Our swimmer sets out from the left
bank swimming at a speed we’ve called 𝑣 with respect to the water, where 𝑣 is
0.500 meters per second. If the river had no current, the
swimmer would end up on the opposite bank directly across from where he started. But there is a current in the river
moving with a speed we’ve called 𝑣 sub 𝑤. This affects the trajectory of the
swimmer so that the actual path he follows is a diagonal line. And he reaches the far shore at
distance we’ve called 𝑑 below the horizontal line of where he started. That distance is 75.0 meters. Knowing all this, we want to solve
for the current speed of the river.
If we were to zoom in on the
swimmer, we would see that there are two velocity vectors affecting the swimmer’s
motion: one we’ve called 𝑣, which is the swimmer’s self-powered motion across the
river; the other is the river current pushing the swimmer downstream. To solve for the speed of the
current, we want to remember an important fact about motion in multiple
directions. And that is that motion in
perpendicular directions is independent. In our case, this means we can
figure out how long it takes the swimmer to cross the river just by looking at the
swimmer’s own velocity 𝑣. That time to cross the river is
unaffected by 𝑣 sub 𝑤 because 𝑣 sub 𝑤 is acting perpendicular to 𝑣.
If we recall that for a constant
speed, the speed is equal to the distance travelled divided by the time it took to
travel that distance, we can write that 𝑣 — the swimmer’s speed — is equal to 𝑤 —
the width of the river — divided by 𝑡 — the time it takes for the swimmer to cross
the river. That means that the time to cross
is equal to 𝑤 divided by 𝑣 or 30.0 meters divided by 0.500 meters per second,
which is equal to 60.0 seconds. That’s how long it takes the
swimmer to reach the other side. We can reuse this expression 𝑣
equals 𝑑 over 𝑡 now to solve for 𝑣 sub 𝑤. 𝑣 sub 𝑤, the speed of the
current, is equal to the distance downstream the swimmer ended up on the opposite
side divided by the total time it took to cross the river. 75.0 meters divided by 60.0 seconds
is equal to 1.25 meters per second. That’s how fast the current in the
river is flowing.
Then, we imagine that a friend of
the swimmer is standing still on shore and watching as the river is crossed. We want to solve for the speed of
the swimmer relative to that stationary friend. If we look back at a diagram of the
velocity vectors acting on the swimmer, we see that we can combine them to form a
net velocity vector, which we’ve called 𝑣 sub 𝑠. It’s the addition of 𝑣 and 𝑣 sub
𝑤 as vectors. 𝑣 sub 𝑠 is a speed. That is, it’s the magnitude of the
addition of 𝑣 and 𝑣 sub 𝑤. So to find 𝑣 sub 𝑠, we’ll take 𝑣
sub 𝑤, square it, add it to 𝑣 squared, and then take the square root of that
sum. When we plug in the given value for
𝑣 and the value for 𝑣 sub 𝑤 we solved for and enter this expression on our
calculator, we find that 𝑣 sub 𝑠, to three significant figures, is 1.35 meters per
second. That’s the speed of the swimmer
under the influence of the river current relative to a stationary observer on
shore.
In our opening example, we talked
about how an airplane pilot will need to pick a bearing based on current weather
conditions. Let’s do an example now that
involves solving for that sort of information.
A small plane can fly at 175
kilometers per hour in still air. The plane flies in a wind that
blows directly out of the west at 36 kilometers per hour. At what angle west of north must
the plane point in order for it to move directly north? How much time is needed for the
plane to reach a point 300 kilometers directly north of its current position?
Let’s call the angle representing
the bearing of the plane 𝜃. We also want to solve for the
amount of time needed for the plane to travel 300 kilometers directly north from
where it currently is. We’ll call this time 𝑡. In this statement, we’re given the
plane’s speed, 175 kilometers per hour in still air. And we’re given the velocity vector
of the wind. We’re told it blows out of the west
at a speed of 36 kilometers per hour. Let’s start on our solution by
drawing a diagram of the plane in travel and this wind affecting its motion.
If we draw in our four compass
directions: north, south, east, and west, we’re told there’s a wind coming out of
the west. We’ve called the speed of that wind
𝑣 sub 𝑤, given as 36 kilometers per hour. We’re told that our plane wants to
be able to fly in a direction due north including the effect of the wind on the
plane’s flight. To do that, the plane will need to
pick its velocity vector. That is, fly in such a direction
that the western component of its velocity vector is counteracted by the eastern
pushing of the wind. When this happens, the plane’s net
motion will be to the north — that is the vector sum of these two velocities. We’re given the magnitude of the
plane’s velocity, its speed which we’ve called 𝑣 sub 𝑝 given as 175 kilometers per
hour.
To solve for 𝜃, we want to
consider the component of the plane’s velocity that will counteract 𝑣 sub 𝑤. We can write this as an
equation. We write that 𝑣 sub 𝑝, the
plane’s speed, multiplied by the sine of the angle 𝜃 is equal to 𝑣 sub 𝑤. This is the condition we enforce so
that the plane’s net motion can be directly north. Dividing both sides by 𝑣 sub 𝑝
and then taking the arcsine of both sides of the equation, we find that 𝜃 is equal
to the inverse sine of the speed of the wind divided by the speed of the plane. When we plug in for those values —
36 kilometers per hour for the wind speed and 175 kilometers per hour for plane
speed — and calculate this expression, we find, to two significant figures, that 𝜃
is 12 degrees. That’s the direction west of north
at which the plane should head so that its net motion is directly north.
Heading this way, we then imagine
that the plane flies a journey of 300 kilometers north. And we want to solve for the time
that this journey would take. To solve for this time, we can
recall that average speed 𝑣 is equal to the distance travelled divided by the time
it takes to travel that distance. We can rearrange this expression so
that it reads 𝑡 is equal to 𝑑 over 𝑣. In our case, 𝑑 is 300 kilometers
and 𝑣 is 𝑣 sub 𝑝, the plane’s speed, multiplied by the cosine of 𝜃. That’s the northerly component of
the plane’s velocity. Plugging in our values for 𝑣 sub
𝑝 and 𝜃, when we calculate this fraction, we find it’s 1.75 hours or to, one
significant figure, 100 minutes. That’s the amount of time that
would take the plane to go 300 kilometers north.
Let’s now summarize what we’ve
learned about velocity vectors.
An object’s velocity vector gives
us two pieces of information. It tells us how fast the object is
moving and in what direction. Second, an object’s speed is equal
to the magnitude of its velocity. Speed, 𝑠, is equal to magnitude of
𝑣 which for a three-dimensional velocity equals the square root of 𝑣 sub 𝑥
squared plus 𝑣 sub 𝑦 squared plus 𝑣 sub 𝑧 squared. And, finally, velocity vectors can
be added graphically, putting vectors tip to tail with one another or algebraically
by adding their numerical components.
